这道题到底怎么写?
最好写写你的步骤,再讲讲你的思路。谢了
推荐回答(4个)
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这个题目似乎能用的条件太少,要找出思路来很困难,但要知道这个题目的答案其实是很容易的,我们只需找一个特殊情况,如上图,当BAD三点共线时,四边形ABCD退化成三角形BCD,显然:因为AB=AC=AD,三角形BCD为等腰直角三角形,
所以:三角形BCD的面积=BD*AC/2=(1/4)BD^2=16
也就是:ABCD的面积=16,这就是本题的答案。严谨地话,本题不能这么解。
但是,以上却为我们解这个题提供了思路---就是要构造一个斜边为BD(等于BD也行)的等腰直角三角形。
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现在我们正式解这个题,
因为AB=AC=AD,所以:过A,C,D三点的园将以A为圆心
所以:角BDC=(1/2)角BAC=45°
现在BD有,45°也有,我们作BF垂直DC,交DC的延长线于F,则:三角形BDF为等腰直角三角形。这样,以BD为斜边的等腰直角三角形就构造好了。
我们只需证明三角形BCF的面积等于三角形ABD的面积即可。
对于这两个三角形,我们现在能有的条件是BD=(根号2)*BF,以及角ABD=角CBF
但一个是直角三角形,一个不是,不好办。而我们有园,所以可以把三角形ABD“拓展”成直角三角形。
延长BA到E,使得AE=AB,那么BE就是园的直径。连接ED,则:角EDB为直角
所以:RT三角形BED相似于RT三角形BCF
所以:三角形BCF的面积/三角形BED的面积=(BF/BD)^2=1/2
即:三角形BCF的面积=(1/2)三角形BED的面积
而:三角形ABD的面积=(1/2)三角形BED的面积
所以:三角形ABD的面积=三角形BCF的面积
所以:四边形ABCD的面积=三角形ABD的面积+三角形BDC的面积
=三角形BCF的面积+三角形BDC的面积=三角形BDF的面积=16
由于用3除余1,所以这个数为3n+1(n为正整数)。 要使3n+1这个数继而满足用5除余2的条件,可用n=1,2,3……来试代,发现当n=2时,3*2+1=7满足条件。 由于15能被3和5整除,所以1...
解:如图:因为AB=AC=AD, 设AB=x, ∆ABC和∆ACD是等腰三角形,因为AB⊥AC,而∆ABC是等腰直角三角形,BC=√2x; 设∠CAD=ß; ∠ABC=∠ACB=a=45º.
BD^2=2x^2-2x^2cos(90º+ß)=2x^2(1+sinß)......(1)
BD^2=CD^2+2x^2-2xCDcos[45º+(180º-ß)/2]=CD^2+2x^2+2xCDsin(45º-ß/2).....(2)
x^2=x^2+CD^2-2xCDcos[(180º-ß)/2]. 即 CD=2xsin(ß/2)......(3)
将(3)代入(2),得:
BD^2=[2xsin(ß/2)]^2+2x^2+2x^2sin(ß/2)sin[ß/2-45º)
=2x^2(2-cosß)-x^2[cos90º-cos(ß-90º)]
=3x^2-2x^2cosß-x^2sinß......(4)
(4)-(1),整理,得: 1-2√(1-sin^2ß)-sinß=0. 即:2√(1-sin^2ß)=1-sinß
方程两边同时平方,得:4sin^2ß=1+sin^2ß-2sinß
3sin^2ß+2sinß-1=0
解得:sinß1=1/3,sinß2=-1(不合理,舍去)
将sinß=1/3代入(1)
x^2=BD^2/[2(1+sinß)]=8^2/[2(1+1/3)]=24.
x=√24=2√6
那么:S=(1/2)(S∆ABC+S∆ACD)=(1/2)(2√6)^2(1+sinß)=12*(1+1/3)=16
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1。三个未知数三个方程,这是确保有解的充要条件。
2. 从未知数来说,三条边相等+一条边+一个角共三个未知数。
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